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### Algebra generated by the set of singletons of an infinite countable set

Note $(A \cup B)'$ is a subset of $B'$. Thus, if $B'$ is finite, then $(A \cup B)'$ is also finite.
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### Comparison between the Lebesgue integral of a given function over two different sets.

No, it is not true. You would additionally need that $m(A) \leq m(B)$. For a counterexample, consider $f \equiv 1$ and $A=[0,1]$, $B=\{2\}$. Edit: with the additional assumption, we can prove the ...
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Start from $$\sum_{\varepsilon,\delta}f_{\varepsilon,\delta}\mu(A_\varepsilon \cap B_\delta)=\sum_{\varepsilon}\sum_{k=1}^m\varepsilon_ka_k \sum_{\delta:A_\varepsilon\cap B_\delta\neq\emptyset}\mu(... 1 vote Accepted ### Simple question that relates Lebesgue measure and Lebesgue integral. It depends on how you defined the Lebesgue integral, but if you started with integrals of simple functions (as in linear combinations of indicator functions of measurable sets), then your equality is ... 3 votes ### Different definitions of the canonical form of simple functions (issue with constant reals different from 0) It seems to me that the first definition adds the restriction that r_k are non-zero to make the decomposition unique. If, say, r_{N+1} is allowed to be 0, one could add it to the sum or discard ... 0 votes ### Different definitions of the canonical form of simple functions (issue with constant reals different from 0) I think that this restriction it for convenience: if you endow the ambient space with some infinite measure, you just have to say that all the E_k must have finite measure for \varphi to be ... 2 votes Accepted ### \| f \|_{L^1} > 0  is stronger than simply f \neq 0. But how much? Yes; if \|f\|_p>0 then f>0 on some positive measure set, since otherwise f=0 a.e. and hence \|f\|_p = 0. Conversely, if \|f\|_p = 0 then f=0 almost everywhere. 1 vote Accepted ### Lebesgue integral of gradient If f is merely continuous you can’t say anything even in the one-dimensional case of the fundamental theorem of calculus, as it may not be differentiable. You need f to be absolutely continuous in ... 1 vote Accepted ### Understanding proof of "if f is non-negative, then it is the supremum of simple functions." For your first question, yes that is true, but I am not sure why that would be helpful in proving \int_B h \, d\mu \le \sup\{\cdots\}. For your second question: This follows from the definition of ... 0 votes ### Prove that a family of functions bounded by a L^1 function is uniformly integrable: \varepsilon-\delta proof Without loss of generality, k = 1. Given \varepsilon>0, choose \delta >0 such that$$\int_A |g|\leq \varepsilon^2,\quad \mu(A)\leq \delta$$Then$$\int_A |f_i|\leq\left\{ \int_Af_i^2\right\...
No, I do not think it is. And if by your observation that $u$ being square integrable implies it being finite almost everywhere you mean that it must be bounded (i.e., $|u(x,t)|\leq k$ for some $k$) ...